A constant 10-newton horizontal force is applied to a 20-kilogram cart at rest on a level floor. If friction is negligible, what is the speed of the cart by the time it has been pushed 8.0 metres?
Let’s call the final speed of the cart 𝑣 sub 𝑓. If we draw a diagram of the cart being pushed, we see that the cart undergoes a horizontal force of 10 newtons and over a span of 8.0 metres picks up speed after it starts from rest. We’ll only consider forces that act in the horizontal direction and modify motion to the right as motion in the positive direction.
Because friction is negligible in this situation, there is a net force that acts on the cart. By Newton’s second law, that means the car will experience an acceleration. And as the force is constant, so is the acceleration. Because of that, the kinematic equations for motion apply to this situation. Scanning through this list, we see that the second equation applies well to our scenario.
In our case, the final velocity 𝑣 sub 𝑓 is what we want to solve for. 𝑣 sub zero is zero because the cart starts from rest. By Newton’s second law, the acceleration 𝑎 is equal to force divided by mass and the distance 𝑑 is given to us; that’s, 8.0 metres. So 𝑣 sub 𝑓 squared is equal to two times the force divided by the mass multiplied by the distance the cart moves.
If we take the square root of both sides of this equation, then the square root and square on the left cancel out leaving us with a simplified expression for 𝑣 sub 𝑓. If we enter in the values for 𝑓, 𝑚, and 𝑑, we find when we compute the square root of two times 10 newtons divided by 20 kilograms multiplied by 8.0 metres a final speed to two significant figures of 2.8 metres per second. That’s how fast the cart is moving after being pushed by a force of 10 newtons for 8.0 metres.